/* 1068:Parencodings */
/*
描述
Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways:
q	By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
q	By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).

Following is an example of the above encodings:

    S		(((()()())))
    P-sequence	    4 5 6666
    W-sequence	    1 1 1456

Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
输入
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is an integer n (1 <= n <= 20), and the second line is the P-sequence of a well-formed string. It contains n positive integers, separated with blanks, representing the P-sequence.
输出
The output file consists of exactly t lines corresponding to test cases. For each test case, the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.
*/



#include <iostream>
#include<cstdio>
#include<string>

using namespace std;

int f[25];
int main()
{
    int t,n,i,sign,p,j;
    string s;
    scanf("%d",&t);
    while(t--)
    {
        sign=0;
        s.clear();
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d",&p);
            while(sign<p)
            {
                sign++;
                s+="(";
            }
            s+=")";
            int q1=1,q2=1;
            for(j=s.length()-2;j>=0&&q2;j--)
            {
                if(s[j]==')')
                {
                    q1++;
                    q2++;
                }
                else q2--;
            }
            f[i]=q1;
        }
        for(i=0;i<n;i++)
            printf("%d%c",f[i],i==n-1?'\n':' ');
    }
    return 0;
}

// Accepted
